How to use the div instruction to find remainder in x86 assembly?

Question

mov eax, 0
mov ebx, 0
mov edx, 0
mov ax, 31
mul cx
mov bx, 12
div bx
add bp, ax
mov eax, 0
mov ebx, 0
mov bp, bp
mov al, 7
div al

can anyone tell me whats wrong with the div al instruction in this block of code, so as I'm debugging every number of bp i calculated, when i divide by al it give me 1 as the remainder, why is this happen?

the remainder should be store back to ah register

thank in advance

edited code :

mov eax, 0
mov ebx, 0
mov edx, 0
mov ax, 31
mul cx
mov bx, 12
div bx
add bp, ax
mov eax, 0
mov ebx, 0
mov ax, bp
mov bl, 7
div bl
mov al, 0

How to without requiring `div`: http://stackoverflow.com/questions/8021772/assembly-language-how-to-do-modulo GCC 4.8 does not use div by default because it is slow: http://stackoverflow.com/questions/4361979/how-does-the-gcc-implementation-of-module-work-and-why-does-it-not-use-the — Ciro Santilli Путлер Капут 六四事, Oct 20 '15 at 20:30

score 4 · Answer 1 · answered May 06 '13 at 17:49

4

You can't use al as divisor, because the command div assumes ax to be the dividend.

This is an example for dividing bp by 7

mov ax,bp // ax is the dividend
mov bl,7  // prepare divisor
div bl    // divide ax by bl

This is 8 bit division, so yes the remainder will be stored in ah. The result is in al.

To clarify: If you write to al you partially overwrite ax!

|31..16|15-8|7-0|
        |AH.|AL.|
        |AX.....|
|EAX............|

answered May 06 '13 at 17:49

typ1232

5,395
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@bluebk where do you get integer overflow? you should not write anything to al if you want to divide bp by something, because you will overwrite ax (the dividend) – typ1232 May 06 '13 at 17:56
i got integer over flow at div bl instruction in the edited code – bluebk May 06 '13 at 17:57
@bluebk well then maybe this is because your result does not fit into `al`. If `bp` is anything greater than 0x700 you will get the integer overflow – typ1232 May 06 '13 at 18:06
my bp for example is 9E8, then should i use bx instead of bl? – bluebk May 06 '13 at 18:08
@bluebk you can't do a 8 bit division of 9b8 by 7. the result is greater than 0xff. – typ1232 May 06 '13 at 18:10

score 1 · Accepted Answer · answered May 06 '13 at 18:20

1

edited code:

mov eax, 0
mov ebx, 0
mov ax, 31
mul cx
mov bx, 12
div bx
add bp, ax
mov eax, 0
mov ebx, 0
mov edx, 0
mov ax, bp
mov bx, 7
div bx
mov esi, edx 
mov eax, 0

answered May 06 '13 at 18:20

bluebk

169
1
6
20

How to use the div instruction to find remainder in x86 assembly?

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